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A body describes simple harmonic motion with an amplitude of 5 cm and a period of `0.2s`. Find the acceleration and velocity of the body when the displacement is (a) 5cm, (b) 3cm, (c) 0 cm. |
Answer» Here, r= 5cm , 0.05 m, T= 0.2 s `omega=(2pi)/T=(2pi)/0.2 =10 pi rad//s` when displacement, is y, then acceleration , `A=-omega^(2) y` velocity , `V=omegasqrt(r^(2)-y^(2))` Case (a) when y=5 cm =0.05 m `A=-(10pi)^(2)xx0.05=-5pi^(2)m//s^(2)` `v=10 pisqrt((0.05)^(2)-(0.05)^(2))=0` Case (b) when `y=3 cm =0.03 m ` `A=-(10pi)^(2)xx0.03=-3pi^(2)m//s^(2)` `V=10pisqrt((0.05)^(2)-(0.03)^(2))=10pixx0.04=0.4 pi m//s` Case (c) when `y=0, A=-(10pi)^(2) xx0=0` `V=10pi sqrt((0.05)^(2)-0^(2))=10pixx0.05 =0.5 pi m//s` |
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